Problem: Let $R$ be the region that is inside the polar curve $r=3$ and outside the polar curve $r=2+\sin(\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2} \int_{0}^{\pi}9\,d\theta+\dfrac{1}{2} \int_{\pi}^{2\pi}\left(2+\sin(\theta)\right)^2\,d\theta$ (Choice B) B $\dfrac12\int_0^{2\pi}\left( 9-\left(2+\sin(\theta)\right)^2\right)d\theta$ (Choice C) C $\dfrac12\int_0^{\pi}\left( 9-\left(2+\sin(\theta)\right)^2\right)d\theta$ (Choice D) D $\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}9\,d\theta+\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\left(2+\sin(\theta)\right)^2\,d\theta$
Since we are dealing with two separate polar curves, a good first step is to identify two areas, each enclosed by a single curve, that together define $R$. Such are $R_1$ and $R_2$ : $y$ $x$ $ R_1$ $ R_2$ $ 1$ $ 1$ $R_1$ is the entire region enclosed by $r=3$ and $R_2$ is the entire region enclosed by $r=2+\sin(\theta)$. Once we express them as integrals, we can find $R$ using the following relationship: $\text{Area of }R=\text{Area of }R_1-\text{Area of }R_2$ $R_1$ is enclosed by $r=3$ between $\alpha=0$ and $\beta=2\pi$ : $\begin{aligned} \text{Area of }R_1&=\dfrac{1}{2} \int_{0}^{2\pi}\left(3\right)^2\,d\theta \\\\ &=\dfrac{1}{2} \int_{0}^{2\pi}9\,d\theta \end{aligned}$ $R_2$ is enclosed by $r=2+\sin(\theta)$ between $\alpha=0$, and $\beta=2\pi$ : $\begin{aligned} \text{Area of }R_2&=\dfrac{1}{2} \int_{0}^{2\pi}\left(2+\sin(\theta)\right)^2\,d\theta \end{aligned}$ Now we can express the area of $R$ : $\begin{aligned} &\phantom{=}\text{Area of }R \\\\ &=\text{Area of }R_1-\text{Area of }R_2 \\\\ &=\dfrac{1}{2} \int_{0}^{2\pi}9\,d\theta-\dfrac{1}{2} \int_{0}^{2\pi}\left(2+\sin(\theta)\right)^2\,d\theta \\\\ &=\dfrac12\int_0^{2\pi}\left( 9-\left(2+\sin(\theta)\right)^2\right)d\theta \end{aligned}$